Solve {l}{14x_{1}+20x_{2}+2x_{3}+49x_{4}=0}{22x_{1}+32x_{2}+3x_{3}+79x_{4}=0}

$$\left\{ \begin{array} { l } { 14 x _ { 1 } + 20 x _ { 2 } + 2 x _ { 3 } + 49 x _ { 4 } = 0 } \\ { 22 x _ { 1 } + 32 x _ { 2 } + 3 x _ { 3 } + 79 x _ { 4 } = 0 } \end{array} \right.$$

Solve for x_1, x_2

$$x_{1}=\frac{3x_{4}-x_{3}}{2}$$

$$x_{2}=\frac{x_{3}}{4}-\frac{7x_{4}}{2}$$

Steps Using Substitution

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$$14x_{1}+20x_{2}+2x_{3}+49x_{4}=0,22x_{1}+32x_{2}+3x_{3}+79x_{4}=0$$

Choose one of the equations and solve it for x_{1} by isolating x_{1} on the left hand side of the equal sign.

$$14x_{1}+20x_{2}+2x_{3}+49x_{4}=0$$

Subtract 2x_{3}+49x_{4} from both sides of the equation.

$$14x_{1}+20x_{2}=-2x_{3}-49x_{4}$$

Subtract 20x_{2} from both sides of the equation.

$$14x_{1}=-20x_{2}-2x_{3}-49x_{4}$$

Divide both sides by 14.

$$x_{1}=\frac{1}{14}\left(-20x_{2}-2x_{3}-49x_{4}\right)$$

Multiply \frac{1}{14} times -20x_{2}-2x_{3}-49x_{4}.

$$x_{1}=-\frac{10}{7}x_{2}-\frac{x_{3}}{7}-\frac{7x_{4}}{2}$$

Substitute -\frac{10x_{2}}{7}-\frac{x_{3}}{7}-\frac{7x_{4}}{2} for x_{1} in the other equation, 22x_{1}+32x_{2}+3x_{3}+79x_{4}=0.

$$22\left(-\frac{10}{7}x_{2}-\frac{x_{3}}{7}-\frac{7x_{4}}{2}\right)+32x_{2}+3x_{3}+79x_{4}=0$$

Multiply 22 times -\frac{10x_{2}}{7}-\frac{x_{3}}{7}-\frac{7x_{4}}{2}.

$$-\frac{220}{7}x_{2}-\frac{22x_{3}}{7}-77x_{4}+32x_{2}+3x_{3}+79x_{4}=0$$

Add -\frac{220x_{2}}{7} to 32x_{2}.

$$\frac{4}{7}x_{2}-\frac{22x_{3}}{7}-77x_{4}+3x_{3}+79x_{4}=0$$

Add -\frac{22x_{3}}{7}-77x_{4} to 3x_{3}+79x_{4}.

$$\frac{4}{7}x_{2}-\frac{x_{3}}{7}+2x_{4}=0$$

Subtract -\frac{x_{3}}{7}+2x_{4} from both sides of the equation.

$$\frac{4}{7}x_{2}=\frac{x_{3}}{7}-2x_{4}$$

Divide both sides of the equation by \frac{4}{7}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$x_{2}=\frac{x_{3}}{4}-\frac{7x_{4}}{2}$$

Substitute \frac{x_{3}}{4}-\frac{7x_{4}}{2} for x_{2} in x_{1}=-\frac{10}{7}x_{2}-\frac{x_{3}}{7}-\frac{7x_{4}}{2}. Because the resulting equation contains only one variable, you can solve for x_{1} directly.

$$x_{1}=-\frac{10}{7}\left(\frac{x_{3}}{4}-\frac{7x_{4}}{2}\right)-\frac{x_{3}}{7}-\frac{7x_{4}}{2}$$

Multiply -\frac{10}{7} times \frac{x_{3}}{4}-\frac{7x_{4}}{2}.

$$x_{1}=-\frac{5x_{3}}{14}+5x_{4}-\frac{x_{3}}{7}-\frac{7x_{4}}{2}$$

Add -\frac{x_{3}}{7}-\frac{7x_{4}}{2} to -\frac{5x_{3}}{14}+5x_{4}.

$$x_{1}=\frac{3x_{4}-x_{3}}{2}$$

The system is now solved.

$$x_{1}=\frac{3x_{4}-x_{3}}{2},x_{2}=\frac{x_{3}}{4}-\frac{7x_{4}}{2}$$

Steps Using Matrices

Put the equations in standard form and then use matrices to solve the system of equations.

$$14x_{1}+20x_{2}+2x_{3}+49x_{4}=0,22x_{1}+32x_{2}+3x_{3}+79x_{4}=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}14&20\\22&32\end{matrix}\right)\left(\begin{matrix}x_{1}\\x_{2}\end{matrix}\right)=\left(\begin{matrix}-2x_{3}-49x_{4}\\-3x_{3}-79x_{4}\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}14&20\\22&32\end{matrix}\right).

$$inverse(\left(\begin{matrix}14&20\\22&32\end{matrix}\right))\left(\begin{matrix}14&20\\22&32\end{matrix}\right)\left(\begin{matrix}x_{1}\\x_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}14&20\\22&32\end{matrix}\right))\left(\begin{matrix}-2x_{3}-49x_{4}\\-3x_{3}-79x_{4}\end{matrix}\right)$$

The product of a matrix and its inverse is the identity matrix.

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x_{1}\\x_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}14&20\\22&32\end{matrix}\right))\left(\begin{matrix}-2x_{3}-49x_{4}\\-3x_{3}-79x_{4}\end{matrix}\right)$$

Multiply the matrices on the left hand side of the equal sign.

$$\left(\begin{matrix}x_{1}\\x_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}14&20\\22&32\end{matrix}\right))\left(\begin{matrix}-2x_{3}-49x_{4}\\-3x_{3}-79x_{4}\end{matrix}\right)$$

For the 2\times 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the inverse matrix is \left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right), so the matrix equation can be rewritten as a matrix multiplication problem.

$$\left(\begin{matrix}x_{1}\\x_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{32}{14\times 32-20\times 22}&-\frac{20}{14\times 32-20\times 22}\\-\frac{22}{14\times 32-20\times 22}&\frac{14}{14\times 32-20\times 22}\end{matrix}\right)\left(\begin{matrix}-2x_{3}-49x_{4}\\-3x_{3}-79x_{4}\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x_{1}\\x_{2}\end{matrix}\right)=\left(\begin{matrix}4&-\frac{5}{2}\\-\frac{11}{4}&\frac{7}{4}\end{matrix}\right)\left(\begin{matrix}-2x_{3}-49x_{4}\\-3x_{3}-79x_{4}\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x_{1}\\x_{2}\end{matrix}\right)=\left(\begin{matrix}4\left(-2x_{3}-49x_{4}\right)-\frac{5}{2}\left(-3x_{3}-79x_{4}\right)\\-\frac{11}{4}\left(-2x_{3}-49x_{4}\right)+\frac{7}{4}\left(-3x_{3}-79x_{4}\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x_{1}\\x_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{3x_{4}-x_{3}}{2}\\\frac{x_{3}}{4}-\frac{7x_{4}}{2}\end{matrix}\right)$$

Extract the matrix elements x_{1} and x_{2}.

$$x_{1}=\frac{3x_{4}-x_{3}}{2},x_{2}=\frac{x_{3}}{4}-\frac{7x_{4}}{2}$$

Steps Using Elimination

In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.

$$14x_{1}+20x_{2}+2x_{3}+49x_{4}=0,22x_{1}+32x_{2}+3x_{3}+79x_{4}=0$$

To make 14x_{1} and 22x_{1} equal, multiply all terms on each side of the first equation by 22 and all terms on each side of the second by 14.

$$22\times 14x_{1}+22\times 20x_{2}+22\left(2x_{3}+49x_{4}\right)=0,14\times 22x_{1}+14\times 32x_{2}+14\left(3x_{3}+79x_{4}\right)=0$$

Simplify.

$$308x_{1}+440x_{2}+44x_{3}+1078x_{4}=0,308x_{1}+448x_{2}+42x_{3}+1106x_{4}=0$$

Subtract 308x_{1}+448x_{2}+42x_{3}+1106x_{4}=0 from 308x_{1}+440x_{2}+44x_{3}+1078x_{4}=0 by subtracting like terms on each side of the equal sign.

$$308x_{1}-308x_{1}+440x_{2}-448x_{2}+44x_{3}+1078x_{4}-42x_{3}-1106x_{4}=0$$

Add 308x_{1} to -308x_{1}. Terms 308x_{1} and -308x_{1} cancel out, leaving an equation with only one variable that can be solved.

$$440x_{2}-448x_{2}+44x_{3}+1078x_{4}-42x_{3}-1106x_{4}=0$$

Add 440x_{2} to -448x_{2}.

$$-8x_{2}+44x_{3}+1078x_{4}-42x_{3}-1106x_{4}=0$$

Add 44x_{3}+1078x_{4} to -42x_{3}-1106x_{4}.

$$-8x_{2}+2x_{3}-28x_{4}=0$$

Subtract 2x_{3}-28x_{4} from both sides of the equation.

$$-8x_{2}=28x_{4}-2x_{3}$$

Divide both sides by -8.

$$x_{2}=\frac{x_{3}}{4}-\frac{7x_{4}}{2}$$

Substitute \frac{x_{3}}{4}-\frac{7x_{4}}{2} for x_{2} in 22x_{1}+32x_{2}+3x_{3}+79x_{4}=0. Because the resulting equation contains only one variable, you can solve for x_{1} directly.

$$22x_{1}+32\left(\frac{x_{3}}{4}-\frac{7x_{4}}{2}\right)+3x_{3}+79x_{4}=0$$

Multiply 32 times \frac{x_{3}}{4}-\frac{7x_{4}}{2}.

$$22x_{1}+8x_{3}-112x_{4}+3x_{3}+79x_{4}=0$$

Add 8x_{3}-112x_{4} to 3x_{3}+79x_{4}.

$$22x_{1}+11x_{3}-33x_{4}=0$$

Subtract 11x_{3}-33x_{4} from both sides of the equation.

$$22x_{1}=33x_{4}-11x_{3}$$

Divide both sides by 22.

$$x_{1}=\frac{3x_{4}-x_{3}}{2}$$

The system is now solved.

$$x_{1}=\frac{3x_{4}-x_{3}}{2},x_{2}=\frac{x_{3}}{4}-\frac{7x_{4}}{2}$$